Sao dài quá z?

280 nhân 35:100=98kg

280 : 100 x 35 = 98 kg

/HT\

TL

là 98 kg 

nha bn

HT

a) Ta có: \(Q=\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)

\(=\dfrac{3x+3\sqrt{x}-3-\left(x-1\right)-\left(x-4\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

b) Thay \(x=4+2\sqrt{3}\) vào Q, ta được:

\(Q=\dfrac{\sqrt{3}+1+1}{\sqrt{3}+1-1}=\dfrac{2+\sqrt{3}}{\sqrt{3}}=\dfrac{2\sqrt{3}+3}{3}\)

c) Để Q=3 thì \(\sqrt{x}+1=3\sqrt{x}-3\)

\(\Leftrightarrow\sqrt{x}-3\sqrt{x}=-3-1\)

\(\Leftrightarrow2\sqrt{x}=4\)

hay x=4

d) Để \(Q>\dfrac{1}{2}\) thì \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{1}{2}>0\)

\(\Leftrightarrow\dfrac{2\sqrt{x}+2-\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}>0\)

\(\Leftrightarrow\sqrt{x}-1>0\)

\(\Leftrightarrow x>1\)

Kết hợp ĐKXĐ, ta được: x>1

e) Để Q nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-1\)

\(\Leftrightarrow2⋮\sqrt{x}-1\)

\(\Leftrightarrow\sqrt{x}-1\in\left\{-1;1;2\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{0;2;3\right\}\)

hay \(x\in\left\{0;4;9\right\}\)

Cơ chế hình thành cây có kiểu gen Aaa là do rồi loạn giảm phân, diễn ra ở kì sau của giảm phân 1. 

Sơ đồ lai: 

P:       Aa          x             Aa

GP: Aa ; 0  ;           A   ;    a

F1:  Aaa             ;            a

lỗi ảnh 

???

Câu 19: B

Câu 20: A

Bài 8: 

a: Ta có: \(\left(5x+1\right)^2=\dfrac{36}{49}\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{-1}{7}\\5x=\dfrac{-13}{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{35}\\x=\dfrac{-13}{35}\end{matrix}\right.\)

b: Ta có: \(\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{2}{3}\right)^6\)

\(\Leftrightarrow x-\dfrac{2}{9}=\dfrac{4}{9}\)

hay \(x=\dfrac{2}{3}\)

\(7,\\ a,=\dfrac{3^{10}\cdot3^5\cdot5^5}{5^6\cdot\left[-\left(3^7\right)\right]}=\dfrac{3^8}{-5}=-\dfrac{6561}{5}\\ b,=8+3-\dfrac{1}{4}\cdot4+\left(4:\dfrac{1}{2}\right)\cdot8\\ =8+3-1+64=74\\ 8,\\ a,\left(5x+1\right)^2=\dfrac{36}{49}\Rightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{35}\\x=-\dfrac{13}{35}\end{matrix}\right.\)

\(b,\Rightarrow8x-1=5\Rightarrow x=\dfrac{3}{4}\\ d,\Rightarrow\left\{{}\begin{matrix}x-3,5=0\\y-\dfrac{1}{10}=0\end{matrix}\right.\left[\left(x-3,5\right)^2\ge0;\left(y-\dfrac{1}{10}\right)^4\ge0\right]\\ \Rightarrow\left\{{}\begin{matrix}x=3,5\\y=\dfrac{1}{10}\end{matrix}\right.\)

a: góc DAC=90-40=50 độ

b: góc ADB=90 độ

c: góc DAB=90-80=10 độ

=>góc BAE=10+50=60 độ

góc AED=180-60=120 độ